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I've got this circuit enter image description here

and I want to find the output voltage \$U_A\$. We are given the values of the resistances, the input \$U_E\$ and the "supplying" Voltage \$U_V\$ (which isn't drawn here):

\$U_E = 10V\$; \$U_V = \pm 15V\$;

\$R_1 = R_2 = 3,9 k\Omega \$; \$R_3 = 8,2 k\Omega \$ ; \$R_4 = 2,2 k\Omega \$

We were supposed to assume that \$U_{E-} = 0\$ and \$ I_{E-} = I_{E+} = 0\$

Attempt at solution:

By KCL I know that: \$ I_1 = I_2 + I_3 \$ and also that \$ I_2 = I_4 \$

By KVL I know that: \$U_E - U_{R1} - U_{R2} = 0\$

and: \$U_E - U_{R_1} - U_{R3} - U_{R4} = 0\$

However, I don't know what happens to \$I_3\$. Is \$I_3\$ just \$I_4\$ or \$I_3 = ?? + I_4\$ and how all of this relates to \$U_A\$ (unless \$U_A = U_{R4}\$ but I am not sure).

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  • \$\begingroup\$ Since U_E+ is grounded the voltage U_A is just the voltage you get across R4 \$\endgroup\$
    – dirac16
    Jan 20, 2017 at 16:05

3 Answers 3

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Here's how I'd do it: -

enter image description here

Once you have found the voltage at this junction (call it Vx) the problem is easy: -

\$U_A = V_x\frac{-R_4}{R_2}\$

How to find Vx?

Maybe convert Ue and Ua to current sources like this: -

enter image description here

Then, the net current flowing through the parallel combination of R1, R2 and R3 is: -

\$\dfrac{U_E}{R_1} - \dfrac{U_A}{R_3}\$

That current multiplied by R1||R2||R3 is Vx

Can you take it from here?

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Here's what you should do: Write node equation for node e and A as below:

enter image description here

  • e: (e-A)/R3 + e/R2 + (e-E)/R1 = 0
  • A: A/R4 + (A-e)/R3 = 0

The variable A stands for the potential U_A. Now you can solve for potential at node A.

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  • \$\begingroup\$ Just that I understand it right: Your equations are \$ I_1 = I_2 + I_3\$ and \$ I_3 + I_4= 0\$ right (expressed in Voltage drop divided by resistance). Thank you very much by the way \$\endgroup\$
    – Eren
    Jan 20, 2017 at 16:42
  • \$\begingroup\$ Yeah,, actually \$\endgroup\$
    – dirac16
    Jan 20, 2017 at 16:53
  • \$\begingroup\$ Alright thanks, thats also the way we were supposed to solve this :) \$\endgroup\$
    – Eren
    Jan 20, 2017 at 17:02
  • \$\begingroup\$ Just a small question: I solved for A now and I always get a Voltage which is below the input Voltage \$U_E\$. Does this make sense in this circumstance? I thought an operational amplifier would amplify the input voltage \$\endgroup\$
    – Eren
    Jan 20, 2017 at 17:33
  • \$\begingroup\$ No, depending on your goal it can do a lot of things. Its not just for amplifying voltages. \$\endgroup\$
    – dirac16
    Jan 20, 2017 at 17:52
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Here comes an alternative method, which seems to be simpler because the whole process is divided in two parts:

1) Ground the opamp output node and find the voltage at the inverting opamp node (forward function Hf) caused by the input VE (see the first diagram).

2) Ground the input node and find the voltage at the inverting opamp node (return functiin Hr) caused by the opamps output voltage VA (see the second diagram).

3.) The ratio (-Hf/Hr) gives the wanted overall transfer function. This simply results from the well-known feedback formula (H. Black) for infinite gain.

schematic

simulate this circuit – Schematic created using CircuitLab

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